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Tunnel-guided air coolers for fresh fruits and vegetables – HidroCooling

Introduction

This fact sheet describes how to design, build and manage a commercial-sized tunnel-guided air cooler for two to six pallets of fresh fruit and vegetables at once. Guided air coolers are commonly used in major areas of growing fresh fruits and vegetables. Figures 1 and 2 show a large guided air cooler used to drupe in a packaging workshop in California. Some crops need to be cooled faster after harvesting than others, so the design is important to ensure that the air flow per unit weight of the product is suitable for the needs of the crops. The principles described in this sheet can be used to support the design of smaller or larger systems as needed. Figure 3 shows a diagram of a guided air cooler.

Figure 1-tunnel-guided air cooler for two-layer pallets of fruit once in a packaging workshop in California. The tunnel is under the black tarpaulin between two rows of pallets.

WHY IS IT NECESSARY TO COOL AS QUICKLY AS POSSIBLE AFTER HARVESTING?

All fresh horticultural crops are living organisms, even after harvesting, and must remain alive and healthy until they are either processed or consumed (Fraser, 1991). The energy needed to continue to live comes from the food reserves in the product itself. The process by which these reserves are converted into energy is called breathing. Thermal energy is released during breathing, but the rate varies depending on the type and variety of the product, the level of maturity, the amount of lesions and the temperature of the product.

Figure 2-a tunnel of pallets of products placed in pairs on a pipe with a fan inside it is formed. A tarp is pulled over the tunnel to force air to circulate past the sides of the container, through the tunnel to the fan.

The temperature of the product has the greatest influence on respiratory activity. Rapid cooling, uniform as quickly as possible after harvesting to eliminate the heat field, is critical in reducing the rate of respiration. This reduces the rate of damage, and provides a longer service life. A general rule is that an hour delay in cooling reduces the shelf life of the product by a day. This is not true for all crops, but it is especially true for very perishable crops during the warm season.

Figure 3-diagram of the back and front of a piping-guided air cooler for fresh products in containers on pallets.

Also, reducing the temperature reduces the rate of ethylene production, loss of moisture, the spread of microorganisms and damage as a result of injuries.

HOW IS GUIDED AIR COOLING PERFORMED?

Air-guided cooling is just one way to quickly remove the heat field from freshly picked products. Most fruits and vegetables can be cooled by air guidance. Large capacity fans are used to draw refrigerated air through the products. It results in rapid and uniform cooling, from high-speed forced convective contact of refrigerated air with hot products. This is different from the cooling chamber, where the product is simply placed in a cold storage chamber and cools slowly and unevenly, mainly through conduction and natural convective contact with refrigerated air.

Pulling air, rather than blowing it, is preferable because it is easier to minimize air short-circuiting and results in much more uniform cooling. Short circuit occurs when air flows go directly to the fan, instead of going through the refrigerated products. Air will not circulate as evenly if pushed the same as when pulled through products. With a proper container design and orientation, products can be quickly and evenly cooled in baskets, boxes, compartments or bags. Air-guided cooling simply does a better job with refrigerated air in cold storage.

Although more expensive, it is better to provide a dedicated room for guided air cooling, and then move the products for long-term storage. Most storage rooms intended for air-guided cooling will have increases in temperature after each fresh batch of hot products added. If this temperature increase is large due to an undersized refrigeration system, cold products that are already in the room may sweat and increase the temperature. Both situations are unacceptable. A good compromise is to form a cooling zone with air guidance by partitioning part of the storage area using a tarp suspended from the ceiling. This helps to reduce temperature fluctuations, but should be considered as a temporary measure.

COOLING TIME 7/8

All fruits and vegetables cool quickly at first, then more slowly over time. Factors affecting the rate of guided air cooling include:

* the density of the product in the container (the less products are placed, the faster they cool down)

* container type, orientation and ventilation characteristics • if air passes evenly and evenly past the product, cooling is faster;

* volume at the surface of the product; the lower the ratio, the faster the cooling (Cherries cool faster than melons);

* the distance traveled by the cooling air (the shorter the distance, the faster the overall cooling is done);

* airflow capacity(the higher the airflow, the faster the cooling).

The relative humidity of the cooling air has a small effect on the loss of moisture, if it is above 85% and the cooling period is less than 1-2 hours.

Regardless of the cooling air temperature or the starting temperature of the product, The Shape of the cooling curve remains the same, provided that all the factors listed above are kept constant. Only the cooling rate changes.

Cooling time 7/8 is an industry standard term describing the time to eliminate the seven-eighth (87.5%) of the temperature difference between the starting temperature of the products and the average cooling temperature (refrigerated air, in the case of air-guided cooling). It is a convenient method of indicating when the products are as close as possible to the cooling temperature. Cooling with air guidance should begin as soon as possible after harvesting, preferably within an hour. Do not allow the products to tighten before placing them in the air-guided cooler, because otherwise they lose their quality and shelf life. The 7/8 cooling time is measured from the moment the product is placed in the air-guided cooler. See Figure 4

Figure 4-typical temperature-time relationship for cooling products.

For example, if a peach at 32oC cooled using 0oc air reaches 4oC in 9 hours, the 7/8 cooling time is 9 hours. That is, a temperature drop of 28oC from 32oC difference between products and air. The 7/8 cooling time is theoretically three times the 1/2 cooling time. So, the same peach for which it took 9 hours to cool to 4oC, would only need 3 hours to reach 16oC, the temperature at 1/2 cooling time, if the rest of the conditions remained the same. In practice, the 7/8 cooling time is usually three times different from the 1/2 cooling time because conditions rarely remain exactly the same during the air-guided cooling period.

Sometimes, one can estimate when a product will be cool 7/8 by knowing other cooling times.

Table 1-relation to cooling times 7/8

For some crops, it may not be necessary to operate the guided air cooler at temperatures as low as optimal for that product. For example, some products could be forced to cool to 5oC, then slowly cooled to room temperature in an adjacent storage room. This compromise could eliminate the need to have a defrost system in the forced air cooling room.

WHAT PRODUCTS CAN BE COOLED BY AIR GUIDANCE?

Most products can be cooled by air guiding, but the 7/8 cooling time needs to be shorter for some products that have special needs:

* have high rates of breathing when harvesting;

* lose moisture slightly (berries/leafy vegetables);

• they are quite mature, like peaches ripe in the tree;

* are shipped to distant markets.

Table 2 lists products requiring rapid cooling, as well as 7/8 cooling times and suggested air flows.

Table 2-relative perishability for fresh fruit and vegetables and cooling times 7/8 times and suggested air flow rates

Crops with very high perishability:

These crops have very high respiration rates at Harvest temperatures and lose moisture quickly after harvesting. They must be quickly cooled immediately after harvesting, or they will have little or no shelf life. Some of these crops are usually Hydro-cooled, frozen or vacuum-cooled. However, they can all be cooled with guided air, cooling is done quickly with high air flows and relatively high humidity to reduce the danger of drying out. It is recommended to use air flows of at least 2 to 6 L/s/kg of products (2 to 6 CFM/lb), trying 7/8 cooling times of no more than 45-90 minutes. These products should be monitored for signs of drying. Sprinkling water before being cooled with guided air (except mushrooms) could be helpful. Do not use the forced air cooler longer than necessary.

High perishable crops:

These crops have high rates of respiration at Harvest temperatures and lose moisture quickly, but it is not so critical to cool as quickly as previously listed. Experience has shown growers that these products should be cooled as quickly as possible after harvesting. Watch for signs of drying of products. It is recommended to use air flows of at least 1 to 3 L/s/kg of Product (1 to 3 CFM/lb), with 7/8 cooling times of no more than 1 to 3 hours.

Fidelity beans should be cooled to about 4oC to 7oC (40oF to 45oF), depending on the grower. Otherwise, they are susceptible to refrigeration damage. Avoid forced air cooling with refrigerated air below 4oC. Try to cool the Fidelity beans in less than 3 hours, if possible. Fidelity beans are often washed after harvesting, so a benefit of forced air cooling is the drying effect of the air flow.

Crops with moderate perishability:

Although these crops are less perishable than those already listed, it is still recommended that these crops be quickly cooled immediately after harvesting. Their validity will be improved. It is recommended to use air flows of at least 0.5 to 1.5 L/s/kg of products (0.5 to 1.5 CFM/lb), and 7/8 cooling times of no more than 3 to 6 hours.

Cantaloupe and summer zucchini are sensitive to refrigeration damage, so forced air cooling with very cold chilled air should be avoided. Cantaloupe should be cooled to about 2oC to 5oC (34oF to 41of), while courgettes should be cooled to about 7oc to 10oC (45oF to 50oF).

WHAT ARE THE COMPONENTS OF THE AIR-GUIDED COOLER

There are four components of an air-guided cooler:

* fan and pipe system

* foam / tarp/plastic to prevent short circuit

* refrigeration system

* monitoring equipment

Fan and pipe system:

The fan feeds the cooling system with air guidance. Air flow is measured in liters of air per second, L/s, (cubes of air per minute or CFM) based on its type (axial or centrifugal); design (blade type and orientation); difficulty in pulling air through products (static pressure); engine size (horsepower or Watts); and revolutions per minute (RPM) of the fan blades.

Fans should be selected based on the air flow they produce at a certain static pressure between the fan Inlet and outlet. For most air-guided cooling systems, static pressures range from about 15 mm to 25 mm (0.6-1.0 inches) of water column. Both centrifugal fans (squirrel cage or furnace type) and axial flow fans (with propellers) can be used. Many air-guided coolers in Canada use centrifugal fans due to the availability of the equipment used and because they are quieter. See Figure 5

Figure 5-centrifugal fan, squirrel cage, or furnace type often used to cool by guiding air.

Many growers find centrifugal fans used for air-guided coolers, but it is difficult to establish air flows for them. However, for planning purposes only, use Table 3 to estimate the capacity of these fans.

Table 3-approximate air flow ranges in L / s (CFM) of centrifugal fans (standard RPM range; one-way input)

Estimating the static pressure against which the air-guided cooling fan must operate is difficult to achieve. Affected by the air flow is the amount of air entering the sides of the container, the alignment of the vent, the distance of air to be traveled to the product, the density of the product in the containers and any restrictions in the air ducts. For most systems, fans should be selected based on a maximum static pressure of 25 mm (1 inch) of water column.

There are several types and models of fans. Smaller centrifugal fans need larger motors running at higher RPM to achieve the same airflow as larger centrifugal fans. In general, larger fans with smaller motors are more efficient. Choose them to reduce operating costs, noise, wear and heat on the refrigeration system to a minimum. This allows the flexibility to install a larger engine in the future to provide higher air flow rates if necessary.

Any air supply area (other than pallets) or return areas (Tunnel) should be designed to keep air speeds below approximately 5 m/s (1000 ft/min). This means providing at least 1 m2 of cross-section for every 5000 L/s of airflow (1 ft21000 CFM). Smaller openings will restrict air flow, make the fan work harder, cause the air to short circuit near the fan, and cause uneven cooling.

To determine the minimum air supply and tunnel cross-section dimensions, suppose an air-guided cooling system is designed to cool 2250 kg (4950lbs) of products on six pallets, with an air flow of 4500 L/s, or 2 L/s per kg (9530 CFM or 1.9 CFM/lb). See Figure 3. Pallets are 1.2 m (4 ft) wide and 1.5 m (5 ft) high. The equation for air flow is:

Q = A X V or a = Q if not then V

where Q is the air flow rate, L / s (CFM)

A, is the cross-sectional area perpendicular to the air flow, M2 (ft2)

V is the speed, m / s (ft / min)

The air flow rate is 4500 L / s (or 4.5 m3/s), so all transverse areas should be minimal:

A = 4.5 m3/s but not 5 m / s = 0.9 m2

(A = 9530 CFM notate notate 1000 ft / min = 9.5 ft2)

So, if the pallets are 1.5 m high( 5 ft), the tunnel should be at least:

W = 0.9 M2 morazo morazo 1.5 m = 0.6 m wide (w = 9.5 ft2 morazo morazo 5 ft = 1.9 ft width)

Estimating the static pressure against which the air-guided cooling fan must operate is difficult to achieve. Affected by the air flow is the amount of air entering the sides of the container, the alignment of the vent, the distance of air to be traveled to the product, the density of the product in the containers and any restrictions in the air ducts. For most systems, fans should be selected based on a maximum static pressure of 25 mm (1 inch) of water column.

There are several types and models of fans. Smaller centrifugal fans need larger motors running at higher RPM to achieve the same airflow as larger centrifugal fans. In general, larger fans with smaller motors are more efficient. Choose them to reduce operating costs, noise, wear and heat on the refrigeration system to a minimum. This allows the flexibility to install a larger engine in the future to provide higher air flow rates if necessary.

For practical reasons, the tunnel should not be less than 0.6 m (2 ft) in width.

The distance of cold air supplied between the outside of the pallets and a wall or pallets of an adjacent air-guided cooler, shall be wide enough to allow cold air to enter easily. For practical reasons, this distance must be at least 0.3 m (1 ft), or more for someone to be able to walk to check. Except in cases where air flows are extremely high, this width is wide enough to allow air to flow freely into the sides of the pallet.

For most applications, the minimum dimensions must be:

* tunnels; 0.6-1.2 m wide • 2-4 ft);

• distance to walls; 0.3-0.6 m width (1-2 ft);

* distance to adjacent units; 0.6-1 m wide (2-3 ft).

Plywood fan casing duct dimensions must be 2.4 M (8 ft) wide and 2.4 m (8 ft) high to support various pallet sizes and heights. It should also be 1.2 m (4 ft) deep from front to back, to help create a more even airflow and to help stabilize the duct, given the weight of the fan on the back of the duct. The opening at the front of the duct for air return to the fan should be centered, 1.2 m (4 ft) wide, and be as high as possible on the duct of the fan housing. See Figure 3.

Most air-guided chillers used in Canada operate with flow rates in the range of 0.5 – 6 L/s/kg of product to be cooled (0.5-6 CFM/lbs). Higher airflow rates can reduce the cooling time, but doubling the airflow does not reduce the time by half. It is very important to understand that higher air flows do not necessarily mean that products will always cool faster, since proper refrigeration and short-circuit prevention are usually more important. It may also be impossible to operate with a very high flow rate, since the fans should be extremely large. There are reports of situations where very high air flows caused static pressures so high that tarps were pulled into the tunnel. No matter how small the flow rates are, any amount of refrigerated air properly drawn through the product will dramatically reduce cooling times compared to a simple cooling chamber.

Foam / tarpaulin / plastic to prevent short circuit

One of the most important, but most often overlooked requirements of a good forced air cooler is the method used to prevent short-circuiting of the cooling air. Air always takes the path of least resistance, so even small cracks need to be clogged. There is no need for a large hole to reduce the flow through the mass of products. A well-designed system can have at least 10% of its air shorted (Thompson, 1996). Poorly designed and operated systems could have most of the air short-circuited.

There are many locations where air can short circuit (see Figure 6):

* openings in the forklift;

* shipping containers that do not fit well on the sides or top or pallet size;

• if the pallets fit on the forced air cooler;

* between the top of the containers on a pallet and a properly unfixed tarpaulin.

Figure 6-clogging locations where air may short-circuit is critical for a cooler with air guiding through the tunnel.

To demonstrate the problem of short circuit, consider the previous example in Figure 6. Cooling air can enter the tunnel only through the outer part of the containers, an area of:

1.5 m x 1.2 m x 3 pallets / part x 2 parts = 10.8 m2

(5 ft x 4 ft x 3 x 2 = 120 ft2)

The required tunnel area was previously calculated to be approximately 1.0 m2 (10.5 ft2). Thus, a short circuit loss area of only 10% would feed all the required air return area. A little air would pass through the products, which already has a higher resistance to air flow. The openings for six-pallet forklifts have a common opening of approx. 0.12 m2 (1.5 ft2) therefore it is important to seal all drain holes.

Heavy plastic or canvas tarps must be pulled over containers with products to help force cooling air to circulate evenly in one direction through products. Heavy strips of foam or door lining are often installed on the front of the duct of the fan housing on which the first pair of pallets are placed, creating an effective air seal. See Figure 3 checking for air leaks after construction is extremely important.

For pallet systems, ideal transport containers for air-guided cooling are those that sit well on all sides and complete the entire footprint of the pallet. Figure 7 compares a straight-walled container that fits on all six sides, up and down like LEGOTM blocks with those that have tapered walls and don’t fit well at the top and bottom. For conical-walled containers, Air short circuits through the conical areas rather than through the product, even if the conical angle is very small (Vigneault & Goyette, 1995). For containers with straight walls that fit well on the sides and top, air must circulate through the product, which results in faster, more uniform cooling. See Figure 8. More precisely, ideal containers for air-guided cooling have vents that represent:

* 25% of the surface perpendicular to the direction of the air flow (Vigneault & Goyette, 1995);

* evenly distributed;

* aligned along the cooling paths;

* long slots rather than round holes to prevent connecting with products;

* no limits imposed by gaskets, trays, packaging materials.

Figure 8-two tunneled forced air coolers, side by side, being used to cool cauliflower in a research test comparing straight-walled containers made of collapsible plastic (left) and conical-walled plastic containers made of embeddable plastic (right). The air will short-circuit through the conical voids. Observe the stiffeners sewn into the tarpaulin to prevent the tarp from being sucked into the tunnel during Operation.

A static pressure indicator of 50 IU or manometer (Figure 9) helps to determine the value of the short circuit. The low pressure tube should be installed inside the tunnel between the pallets as far as possible from the fan (Boyette, 1994). See Figure 3. The high-pressure tube should be installed in the normal air flow of the cold storage room. For most applications, the difference should measure approximately 12 mm (0.5 in.) static column water pressure. This measures the load against which the fan must work to draw air. By sealing the short-circuit gaps, the static pressure will increase, indicating that the fan works harder to draw air through the products and ensures the circulation of cooling air through the containers, and not around them.

Figure 9-a static pressure gauge or indicator can help find air short-circuit gaps.

Common methods of preventing short circuits are: foam seals or doors between pallets and the cooling unit by forcing air; embossed cardboard or plastic strips between pallets and at their ends or at the openings of the forklifts, which are sucked into place by air pressure; or floor dampers on which the pallets rest to prevent short circuits through the openings of the forklifts.

Refrigeration system:

There is an old proverb that says You can never have too much refrigeration in a cold storage room. This certainly applies to cooling systems with air guidance. Since cooling begins immediately after the products are put on the unit, and the slope of the cooling curve is so steep initially (Figure 4), the amount of refrigeration required at the beginning of the cooling process is enormous. This is often much higher than manufacturers can afford or need. The formula for refrigeration in kilowatts, kW, (BTU / h) required at any time is adapted from the formula for momentary cooling rate (Mitchell et al, 1972):

– kW required (BTU/hr) = 2.1 x (A-B) X C X D and E a = product temperature, oC (oF);

– B = cooling air temperature, oC (oF);

– C = weight of the cooling product, kg (lbs).

– D = the specific heat of the products, usually about 3.77 kJ / kg / oC (0.9 Btu/lb/ oF)

– E = cooling time 7/8

In the previous example, the cooling capacity required to cool 2275 kg (5,000 lbs) of strawberries from 28oc (82oF) to 3.5 oC (38oF) in 2 hours, using 0oc (32of) air cooling (7/8 cooling time of 2 hours)?

Using the above formula, momentary refrigeration at the beginning of the cooling process (worst case scenario) would be:

– 2.1 x (28oc – 0oC) X 2275 kg x 3.77 kJ/kg/oC 2 hr = 252,150 kJ/hr or 70 kJ / s or 70 kW

– 2.1 x (82oF-32of) x 5000 lbs x 0.9 Btu / lb./ of Part 2 hr= 236,250 Btu / hr

That’s almost 20 tons of refrigeration! There is 3.5 kW (12000 BTU / h) in a ton of refrigeration, a term used by the industry. Most growers can not afford to design for the worst case. However, if you learn to accept that the room temperature will rise slightly, initially, when the products are placed in the guided air cooler, but that it will gradually recover, you can design with lower levels of refrigeration. With good management, the proposed general rule is to design for about 2/3 of the maximum momentary refrigeration rate at the beginning of the cooling process:

70 kW x 2/3 = 47 kW

(236,250 Btu / hr x 2/3 = 157,500 BTU / hr)

This means 13 tons of refrigeration well above the amount required for thermal loads produced elsewhere in storage, such as through doors, walls and ceilings. The thermal load of the products would probably represent at least 80% of the total heat load in the warehouse.

If the system is not designed for this, do not direct heated air from the forced air cooling fan directly into the evaporator coils or cold air from the evaporator coils directly to the pallets that are cooled with guided air. In most cases, evaporator coils and fans were not designed for this application. When using a forced air cooling system in a room used to store products that are already cooled, direct the exhaust air from the forced air cooling fan away from the products, and towards the evaporator coils.

The relative humidity of the cooling air in the forced air cooling chamber should be more than 85% to prevent the product from wilting. This means large cooling areas of the evaporator coil, and the temperature drops through the cooling coils. If a cold storage room is maintained at 0 o C (32 O F), and the evaporator cooling coils are too small in size, the air coming from the coils will be a few degrees below the freezing point. It dries the air and keeps the relative humidity in the room too low for fresh fruits and vegetables. The products could be damaged by refrigeration if the air is not allowed to heat up slightly in the room at first, before being pulled through the products by the forced air cooler.

It is important to keep the cooling air as close as possible to the set temperature, especially near the end of the forced air cooling period. If the air rises by a few degrees, the product can stop cooling and increase its temperature. This emphasizes the need to have separate cooling chambers for sufficient refrigeration capacity.

Some refrigeration systems, such as the Filacell system, are designed specifically for forced air cooling. It has high capacity fans that can handle high static pressures while providing very high humidity. Consult a refrigeration contractor about the available options.

Monitoring equipment:

Monitoring equipment can help administer the forced air cooler. Some of the most important administration information are:

* Product starting temperature;

* Desired final temperature of the product;

* maximum time the product can be cooled by air guidance.

All these aspects are more important for novice users of a forced air cooler.

The internal temperature should be taken for a few pieces of products from a pallet before placing in the forced air cooler. This means probing the center of the product with good temperature measurement equipment that provides an instant digital readout. See Figure 10. The temperature of the product may not be the same as the temperature of the surrounding air. Large products such as cantaloupe or cabbage will require more to heat (or cool) than smaller products such as plums, even if the ambient air temperature is rising (or falling) rapidly. For example, the temperature of the outdoor air in the middle of the morning can be higher than the temperature of peaches that are still in the tree, because they can still be cold as a result of the previous night or due to shading of the leaves. In contrast, strawberries can be warmer than the air temperature if the sun shines directly on their dark coloratura. Also, products at the top of a container, basket or box may be warmer than those below due to direct sunlight, heat, or heat transfer from a hot, dark container.

Figure 10-a portable, digital temperature reading probe is an essential element

Most operators know at what temperature they want to store their products. Unfortunately, when things get crowded when harvesting, products can sometimes not sit on the forced air cooler as long as necessary. However, by knowing the starting temperature of the product, operators can determine what temperature the product will have after a period of time on the forced air cooler.

It is difficult and time-consuming to monitor the temperature of the product as it cools. However, one method of estimating the actual temperature of the product at any time is to monitor the temperature of the air discharged from the forced air cooling fan, and then compare it with the temperature of the cooling air in the room as it enters the pallet. The exhaust air will be approximately halfway between the difference between the cooling air entering the pallet and the temperature of the products at that time.

If the cooling air in the chamber entering the pallet is 2 oC, and the air discharged from the forced air cooling fan is at 10o C, the product would be at about 18oC, because 10oC is halfway between 2oC and 18oC. Products that receive cold air first will cool down faster than those that are downstream, because downstream products receive warmer air. In the event that there are several air short circuits, this monitoring method is unreliable, since more cold air will reach the fan, lowering the temperature of the exhaust air and giving the operator a false sense of cooling progress.

A thermostat may be placed in the air discharged from the forced air cooling fan for the purpose of shutting it down when the air flow reaches a certain temperature or slowing it down if it is a variable speed fan. This can help prevent the operation of the equipment longer than necessary, save electricity and prevent unnecessary addition of heat from Motors to the cold storage room. A timer to turn off the fan after a period of time could be installed, if any.

10 STEPS TO DESIGN A GUIDED AIR COOLER

1. Determination of average production per day, kg (lbs)

2. Determination of maximum production per day, kg (lbs)

3. Determination of available cooling time (hours / day)

4. Setting the number of lots ( lots / day)

5. Batch size calculation, kg / lot (lbs / lot)

6. Choosing an air flow rate, l / s / kg (CFM / lbs)

7. Calculation of fan airflow, L / s (CFM)

8. Peak refrigeration calculation, kW (Btu / h)

9. Use of refrigeration rule 2/3, KW (Btu / h)

10. Determination of tunnel width and Wall distance, m (ft)

1. Determination of average production per day, kg (lbs)

4 ha x 3000 units / ha x 6 kg / unit÷ 25 d = 2880 kg / d

10 ac x 1200 units / ac x 13.2 lbs / unit÷ 25 d = 6340 lbs / d

2. Determination of maximum production per day, kg (lbs)

The Daily Harvest could range up to 10000 kg of harvested fruit (22000 lbs). It is unrealistic to design for the busiest day of the season, but a general rule of thumb is to design for a busy day that is often twice the average day. So:

2880 kg/average per day x 2 = 5760 kg/usual busy day

(6340 lbs/average per day x 2 = 12680 lbs/usual busy day)

3. Determination of available cooling time (hours / day)

Harvesting is from 6:00 to 12: 00 or 6 hours. The most recently picked fruits are put on the forced air cooler around approx. 7 in the morning, and the fruits come to storage continuously after that until 12:00. Forced air cooling can continue as long as necessary after 12:00, so the available cooling time of 6 hours is estimated , starting at 07: 00

until 1:00. The last fruits picked are generally the warmest fruits picked, so they can sit on the forced air cooler longer if necessary.

4. Setting the number of lots ( lots / day)

From Table 2, it is reasonable to want a 7/8 cooling time

1.5 hours for strawberries, therefore:

6 hours available / day ÷ 1,5 hours / lot = 4 lots / day

5. Calculate lot size, kg / lot (lbs / lot) 5760 kg / day ÷ 4 lots / day = 1440 kg / lot (12680 lbs / Day ÷ 4 lots / day = 3170 lbs / lot )

It would result in 240 units / lot, or 4 pallets.

6. Choosing an air flow rate, l / s / kg (CFM / lbs)

From Table 2, a 7/8 cooling time of 1,5 hours corresponds to an approximate airflow rate of 2,0 l / s / kg of products (2 CFM / lbs). The higher the airflow, the faster the cooling time, and the lower the airflow, the slower the cooling time. Predicting the 7/8 cooling time is difficult because it depends on a lot of variables.

7. Calculation of fan airflow, L / s (CFM)

2.0 L / s / kg x 1440 kg / lot = 2,880 l / s (2,88 m 3 / s)

(2.0 CFM / lbs x 3170 lbs / lot = 6340 CFM)

Table 3 suggests that a centrifugal fan with a 2.25 kW (3 hp) motor would suffice. If not, contact a fan supplier for a fan that distributes at least 2880 l / s at a static pressure of 25 mm (6340 CFM at a static pressure of 1 inch).

8. Peak refrigeration calculation, kW (Btu / h)

2.1 x (25oC – 0oC) x 1440 kg x 3.77 kJ/kg/oC÷1.5 h

= 190000 kJ / h or 53 kJ / s or 53 kW or 15 tonnes

2.1 x (77oF – 32of) x 3170 lbs x 0.9 Btu/lb/ of ÷1.5 h

= 179750 Btu / h or 15 tonnes of refrigeration

This amount is for cooling berries, not the hall itself!

9. Use of refrigeration rule 2/3, KW (Btu / h)

15 tons of refrigeration is a lot for a forced air cooler with 4 pallets of berries at a time. So:

53 kW (theory) x 2/3 = 35 kW (practical); 10 tonnes

179750 Btu / h x 2/3 = 119,800 Btu / h;10 tonnes

10. Determination of tunnel width and Wall distance, m (ft)

2.88 m3/s ÷ 5 m/S Max. of air = 0.58 M2 surface min.

6340 CFM ÷ 1,000 ft / min max. = 6.34 ft 2 min.

With pallets 1.5 m high( 5 ft), the tunnel width must be at least 0.58 m2 ÷ 1.5 m = 0.4 m, however, a practical minimum size is 0.6 m (2 ft). The gap to the wall would also be at least 0.3 m (1 ft) to allow an operator to sneak in.

OTHER CONSIDERATIONS

* the tarpaulin over the tunnel should extend as close as possible to the upper outer edge of the pallets and up to the floor at the end of the row of pallets to prevent air short-circuiting

* stiffening elements woven into the tarpaulin are necessary to prevent it from being sucked into the tunnel

* check for air leaks using cellophane, which will be pulled into uncoated holes

* the fan should be centered in the duct so that it draws air as evenly as possible from the tunnel

* large lots harvested early in the day have a similar thermal load as small lots harvested later in the same day

* empty containers and harvested products should be covered with tents or canopies in the field to minimize heat accumulation

• as the day warms up, reduce the time the product is in the field before it is cooled

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